## 2018年1月19日金曜日

### 数学 - Python - 解析学 - 多変数の関数 - 微分可能性と勾配ベクトル(偏微分、有界、連続性、単位ベクトル)

1. $\begin{array}{}{D}_{1}f\left(0,0\right)=1\\ \left(x,y\right)\ne 0\\ {D}_{2}f\left(x,y\right)\\ =\frac{3{x}^{2}\left({x}^{2}+{y}^{2}\right)-{x}^{3}·2x}{{\left({x}^{2}+{y}^{2}\right)}^{2}}\\ =\frac{{x}^{4}+3{x}^{2}{y}^{2}}{{\left({x}^{2}+{y}^{2}\right)}^{2}}\\ \le \frac{2{x}^{4}+4{x}^{2}{y}^{2}+2{y}^{2}}{{\left({x}^{2}+{y}^{2}\right)}^{2}}\\ =\frac{2{\left({x}^{2}+{y}^{2}\right)}^{2}}{{\left({x}^{2}+{y}^{2}\right)}^{2}}\\ =2\\ \left|{D}_{1}f\left(x,y\right)\right|\le 2\end{array}$
$\begin{array}{}{D}_{2}f\left(0,0\right)=0\\ \left(x,y\right)\ne 0\\ {D}_{2}f\left(x,y\right)\\ =\frac{-{x}^{3}·2y}{{\left({x}^{2}+{y}^{2}\right)}^{2}}\\ \left|{D}_{2}f\left(x,y\right)\right|\le 2\end{array}$

2. $\begin{array}{}u=\left({u}_{1},{u}_{2}\right)\\ \underset{h\to 0}{\mathrm{lim}}\frac{f\left(h{u}_{1},h{u}_{2}\right)-f\left(0,0\right)}{h}\\ =\underset{h\to 0}{\mathrm{lim}}\frac{\frac{{\left(h{u}_{1}\right)}^{3}}{{\left(h{u}_{1}\right)}^{2}+{\left(h{u}_{2}\right)}^{2}}}{h}\\ =\underset{h\to 0}{\mathrm{lim}}\frac{h{u}_{1}^{3}}{\left({u}_{1}^{2}+{u}_{2}^{2}\right)h}\\ =\underset{h\to 0}{\mathrm{lim}}\frac{{u}_{1}^{3}}{{\left|u\right|}^{2}}\\ ={u}_{1}^{3}\\ {D}_{u}f\left(0,0\right)={u}_{1}^{3}\\ \mathrm{max}\left|{D}_{u}f\left(0,0\right)\right|=1\end{array}$

3. （0，0） において微分可能であると仮定する。

${D}_{n}f\left(0,0\right)=grad\left(f\left(0,0\right)\right)·u=\left(1,0\right)·\left({u}_{1},{u}_{2}\right)={u}_{1}\ne {u}_{1}^{3}\left({u}_{1}\ne 1,0\right)$

よって矛盾。

ゆえに、原点において f は微分可能ではない。

コード(Emacs)

Python 3

#!/usr/bin/env python3
from sympy import pprint, symbols, Matrix, Derivative

x, y = symbols('x, y')
xs = [x, y]
f = x ** 3 / (x ** 2 + y ** 2)
gradf = Matrix([Derivative(f, xi, 1).doit() for xi in xs])

$./sample12.py 3 x ─────── 2 2 x + y ⎡ 4 2 ⎤ ⎢ 2⋅x 3⋅x ⎥ ⎢- ────────── + ───────⎥ ⎢ 2 2 2⎥ ⎢ ⎛ 2 2⎞ x + y ⎥ ⎢ ⎝x + y ⎠ ⎥ ⎢ ⎥ ⎢ 3 ⎥ ⎢ -2⋅x ⋅y ⎥ ⎢ ────────── ⎥ ⎢ 2 ⎥ ⎢ ⎛ 2 2⎞ ⎥ ⎣ ⎝x + y ⎠ ⎦$